The growth of charge Q on a capacitor connected in series with a resistance R ohm and a direct voltage source of emf V0 volt is given by
Q = Q0(1– e–t/RC) where Q0 = CV0 which is the final (maximum) charge on the capacitor and e is the base of natural logarithms.
Q = Q0e–t/RC
The nature of the growth and the decay of charge are shown in the adjoining figure.
Since the voltage (V) across the capacitor is directly proportional to the charge on it, the
V = V0(1– e–t/RC) and
V = V0e–t/RC where V0 is the charging battery
The time constant of the CR circuit is RC which is the time taken for the charge to grow from zero value to (1– 1/e) times (0.632 times) the final maximum charge. This follows from the equation for the exponential growth of charge.
The charging current is obtained by differentiating the equation Q = Q0(1– e–t/RC) with respect to the time t. Thus the charging current (dQ/dt) is given by
I = (Q0 /RC) e–t/RC = I0 e–t/RC where I0 = Q0 /RC = V0/R where V0 is the charging
The exponential decay of the charging current is shown in the adjoining figure.
The current during the discharge of the capacitor through the resistor is obtained by differentiating the equation Q = Q0e–t/RC with respect to the time t. Thus the discharging current (dQ/dt) is given by
I = – (Q0 /RC) e–t/RC = –I0 e–t/RC where I0 = Q0 /RC = V0/R where V0 is the initial voltage on the capacitor.
The exponential decay of the discharging current which flows in the direction opposite to the charging current also is shown in the adjoining figure.
[The charging and discharging of a capacitor is similar to inflating and deflating a balloon. The air pressure is the analogue of
We will now discuss a few multiple choice questions in this section:
(1) A transistor time delay relay circuit makes use of a series RC circuit. If the circuit uses a 60 μF capacitor what is the resistance required to produce a
(a) 60 Ω
(b) 6×10–5 Ω
(c) 1.67×104 Ω
(d) 105 Ω
(e) 106 Ω
R = 60 s/60×10–6 F = 106 Ω
(2) A constant current source charges an initially uncharged capacitor. During the charging process, the potential difference V between the plates of the capacitor is related to the charging time t as
(a) V α t
(b) V α t0
(c) V α t–1
(d) V α (1– e–t)
(e) V α e–t
Since the current I is constant, the charge
(3) A 10 μF capacitor is connected in series with a 5 Ω resistor and a battery of negligible internal resistance and emf 6 V. After a long time, if the charging battery is removed and the capacitor is immediately connected across the 5 Ω resistor, the initial value of the discharge current is
(a) 1.2 A
(b) 0.6 A
(c) 0.3 A
(d) 0.83 A
The capacitor will get charged to the battery
I = –I0 e–t/RC where I0 = V0/R where V0 is the initial voltage on the capacitor. Let us ignore the negative sign which just indicates that the discharge current is opposite to the charging current.
Therefore, I = (6/5) e0 = 1.2 A
(4) In the above question what will be the
(a) 1.333 V
(b) 2.208 V
(c) 3.792 V
(d) 4.5 V
(e) 6 V
The time constant of the circuit is RC =5×10×10–6 = 50 microsecond.
After one time constant, the voltage on the capacitor will be (1– 1/e) times (0.632 times) the charging battery
(5) In question No.3 what will be the charging current at the instant t = 50 microseconds?
(a) 0.3680 A
(b) 0.4416 A
(c) 0.6 A
(d) 0.7326 A
(e) 1.2 A
Since the time constant of the circuit, RC =5×10×10–6 s = 50 microseconds, the charging current at the instant t = 50 microseconds will be I = I0 e–t/RC = I0 e–1 = 0.368 I0.
But I0 = V0/R = 6/5 = 1.2 A.
Therefore I = 0.368×1.2 = 0.4416 A.